nNa2O=m/M=15,5/62=0,25 (mol)

PT: Na2O + H2O -> 2NaOH

cứ -: 1................1................2 (mol)

Vậy: 0,25 ------------------->0,5(mol)

=> CM NaOH=n/V=0,5/0,5 =1 (M)

b) Ta có PT:

NaOH + H2SO4 -> Na2SO4 + H2O

1.................1................1...............1 (mol)

0,5 ---------->0,5------->0,5 (mol)

=> mH2SO4=n.M=0,5.98=49(gam)

=> md d H2SO4= mH2SO4.100%C%=49.10020=245(g)

=> Vd d H2SO4=md d H2SO4 / D = 245/1,24≈197,6(ml)≈197,6(ml)=0,1976 lít

Ta có: Vd d sau phản ứng = Vd d H2SO4=0,1976 (lít)

CM=n/M=0,5/0,1976≈2,53(M)